Simple Harmonic Motion

For a two-body system, the angular frequency ( $\omega$ ) is given by :

$ \omega=\sqrt{\frac{\mathrm{k}}{\mu}} $

Where :

• k is the spring constant.

• $\mu$ is the reduced mass of the system.

The reduced mass is calculated using the formula :

$ \mu=\frac{m_1 \cdot m_2}{m_1+m_2} $

Given values of masses are :

• $\mathrm{m}_1=1 \mathrm{~kg}$

• $\mathrm{m}_2=200 \mathrm{~g}=0.2 \mathrm{~kg}$

$ \mu=\frac{1 \times 0.2}{1+0.2}=\frac{0.2}{1.2}=\frac{1}{6} \mathrm{~kg} $

So the angular frequency is,

$ \omega=\sqrt{\frac{150}{1 / 6}} $

$\Rightarrow $ $ \omega=\sqrt{900}=30 \mathrm{rad} / \mathrm{s} $

Therefore, the angular frequency of the system in SI units is 30 . Hence, the correct option is (C).

Initially, the block is floating. The weight of the block is balanced by the buoyant force $\left(\mathrm{F}_{\mathrm{B}}\right)$.

Let h be the height of the submerged part of the block.

$ \mathrm{Mg}=\mathrm{F}_{\mathrm{B}}=\mathrm{V}_{\text {submerged }} \times \rho_{\text {liquid }} \times \mathrm{g} $

$\Rightarrow $ $\mathrm{Mg}=(\mathrm{A} \times \mathrm{h}) \rho \mathrm{g}$

When the block is depressed by a small distance x downwards, the volume of the displaced liquid increases by $A \cdot x$. This creates an extra buoyant force which acts as the restoring force.

The additional buoyant force is :

$ \mathrm{F}_{\text {restoring }}=-(\text { Additional Volume }) \times \rho \times \mathrm{g} $

$ \mathrm{F}_{\text {restoring }}=-(\mathrm{Ax}) \rho \mathrm{g} $

Since the force is proportional to displacement x and acts opposite to it ( $\mathrm{F} \propto-\mathrm{x}$ ), the motion is Simple Harmonic Motion (SHM).

Using Newton's Second Law ( $\mathrm{F}=\mathrm{Ma}$ ) :

$ \mathrm{Ma}=-(\rho \mathrm{Ag}) \mathrm{x} $

$\Rightarrow $ $ a=-(\rho A g M) x . $

Standard equation for SHM acceleration is $\mathrm{a}=-\omega^2 \mathrm{x}$

Comparing the two equations:

$ \omega^2=\frac{\rho \mathrm{Ag}}{\mathrm{M}} \Rightarrow \omega=\sqrt{\frac{\rho \mathrm{Ag}}{\mathrm{M}}} $

The time period T is given by $\mathrm{T}=\frac{2 \pi}{\omega}$.

$\Rightarrow $ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\rho \mathrm{Ag}}}$

Hence, the correct option is (C).

The spring constant k of a spring is inversely proportional to the length L of the spring.

$ \mathrm{k} \propto \frac{1}{\mathrm{~L}} $

This means that the product of the spring constant and the length is a constant for a given spring:

$ \mathrm{k} \cdot \mathrm{~L}=\text { Constant } $

The force constant of original spring is $\mathrm{k}_{\mathrm{o}}=15 \mathrm{~N} / \mathrm{m}$.

Let the original length be L .

The spring is cut into two pieces with a length ratio of $1: 3$.

$ x+3 x=L \Rightarrow x=\frac{L}{4} $

Hence, the length of smaller spring is $L_1=x=\frac{L}{4}$ and the length of larger spring is $L_2=3 x=\frac{3 L}{4}$.

Let $\mathrm{k}_1$ be the force constant of the smaller piece.

Using the inverse proportionality relationship:

$ \mathrm{k}_{\mathrm{o}} \mathrm{~L}_{\mathrm{o}}=\mathrm{k}_1 \mathrm{~L}_1 $

$\Rightarrow $ $ 15 \times \mathrm{L}=\mathrm{k}_1 \times\left(\frac{\mathrm{L}}{4}\right) \Rightarrow \mathrm{k}_1=60 \mathrm{~N} / \mathrm{m} $

Therefore, the force constant of smaller piece is $60 \mathrm{~N} / \mathrm{m}$.

Hence, the correct option is (C).

The time period (T) of a simple pendulum is the time taken to complete one full oscillation.

For a simple pendulum of the length of the string $(\mathrm{L})$ and the acceleration due to gravity $(\mathrm{g})$ the formula for the time period is,

$ \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}} $

So, the time period is directly proportional to the square root of the length $(T \propto \sqrt{L})$.

$ \mathrm{T}^2 \propto \mathrm{~L} \Rightarrow \propto \mathrm{~T}^2 $

Case - 1

The initial length of the string is $\mathrm{L}_1=30 \mathrm{~cm}$

$\mathrm{N}_1=20$ is the number of oscillations completed in total time $\mathrm{t}_1=10 \mathrm{~s}$

So, the initial time period of oscillation is

$ \mathrm{T}_1=\frac{\text { Total Time }}{\text { Number of Oscillations }}=\frac{10}{20}=0.5 \mathrm{~s} $

Case - 2

$\mathrm{N}_2=40$ is the number of oscillations completed in the same time $\mathrm{t}_2=\mathrm{t}_1=10 \mathrm{~s}$

So, the final time period of oscillation is,

$ \mathrm{T}_2=\frac{\text { Total Time }}{\text { Number of Oscillations }}=\frac{10}{40}=0.25 \mathrm{~s} $

We need to find the required length $\left(\mathrm{L}_2\right)$ corresponding to the final length of the string, Using the proportionality $\mathrm{L} \propto \mathrm{T}^2$

$ \frac{\mathrm{L}_2}{\mathrm{~L}_1}=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2 $

$\Rightarrow $ $\frac{\mathrm{L}_2}{30}=\left(\frac{0.25}{0.50}\right)^2=\frac{1}{4}$

$\Rightarrow $ $ \frac{\mathrm{L}_2}{30}=\frac{1}{4} \Rightarrow \mathrm{~L}_2=\frac{30}{4} \mathrm{~cm}=7.5 \mathrm{~cm} $

Therefore, the required length of the string is 7.5 cm .

Hence, the correct option is (B).

The time period $(T)$ of a simple pendulum of length $L$ is given by the standard formula :

$ \mathrm{T}=2 \pi \sqrt{\mathrm{~L} / \mathrm{g}} $

Where $g$ is the acceleration due to gravity.

The graphs in the options plot $\frac{1}{\mathrm{~T}^2}$ versus L .

On rearranging the formula for the time period,

$ \mathrm{T}^2=\frac{4 \pi^2 \mathrm{~L}}{\mathrm{~g}} $

$\Rightarrow $ $\frac{1}{\mathrm{~T}^2}=\frac{\mathrm{g}}{4 \pi^2 \mathrm{~L}}$

$\Rightarrow $$ \frac{1}{\mathrm{~T}^2}=\left(\frac{\mathrm{g}}{4 \pi^2}\right) \frac{1}{\mathrm{~L}} $

Since $g$ and $4 \pi^2$ are constants, the term $\frac{g}{4 \pi^2}$ is just a constant value. Let $\frac{g}{4 \pi^2}=k$, where $k$ is any positive constant.

Therefore, the relationship is :

$ \mathrm{y}=\mathrm{kx} $

where $\mathrm{y}=\frac{1}{\mathrm{~T}^2}$ and $\mathrm{x}=\mathrm{L}$.

The mathematical equation $\mathrm{y}=\mathrm{kx}$ (or $\mathrm{xy}=$ constant) represents a rectangular hyperbola, where $\frac{1}{\mathrm{~T}^2} \rightarrow 0$ if $\mathrm{L} \rightarrow \infty$, i.e., x -axis is horizontal asymptote. Also, $\frac{1}{\mathrm{~T}^2} \rightarrow \infty$ if $\mathrm{L} \rightarrow 0$, i.e., y-axis is horizontal asymptote

So, the resulting graph is,

Therefore, the correct option is (A).

Let the displacement $(\mathrm{x})$ of a simple harmonic oscillator at any time t be represented by :

$ x=A \sin (\omega t) $

Where :

• A is the amplitude.

• $\omega$ is the angular frequency of the oscillator.

The velocity ( $v$ ) is the derivative of displacement with respect to time :

$ v=\frac{d x}{d t}=A \omega \cos (\omega t) $

The kinetic energy (K.E.) of the oscillator is given by :

$ \text { K. } \mathrm{E}=\frac{1}{2} \mathrm{~m} \mathrm{v}^2 $

$\Rightarrow $ K. $\mathrm{E} .=\frac{1}{2} \mathrm{~m}(\mathrm{~A} \omega \cos (\omega \mathrm{t}))^2$

$\Rightarrow $ K. $\mathrm{E} .=\frac{1}{2} \mathrm{~m} \mathrm{~A}^2 \omega^2 \cos ^2(\omega \mathrm{t})$

$\Rightarrow $ K. $E \cdot=\frac{1}{2} m A^2 \omega^2\left[\frac{1+\cos (2 \omega t)}{2}\right]$

$\Rightarrow $ K. $E \cdot=\frac{1}{4} m A^2 \omega^2+\frac{1}{4} m A^2 \omega^2 \cos (2 \omega t)$

$\Rightarrow $ $ \text { K. } \mathrm{E} .=\frac{1}{4} \mathrm{~m} \mathrm{~A}^2 \omega^2+\frac{1}{4} \mathrm{~m} \mathrm{~A}^2 \omega^2 \cos \left(\omega^{\prime} \mathrm{t}\right) $

Here, $\omega^{\prime}=2 \omega$ is the frequency of oscillation of kinetic energy of oscillator.

So, the kinetic energy oscillates with an angular frequency of $2 \omega$.

The kinetic energy oscillates with an angular frequency $\omega^{\prime}=176 \mathrm{rad} / \mathrm{s}$.

$ \Rightarrow \omega=\frac{176}{2}=88 \mathrm{rad} / \mathrm{s} $

Using the relationship between angular frequency ( $\omega$ ) and frequency (f) in Hz is:

$ \omega=2 \pi \mathrm{f} $

$\Rightarrow $ $ f=\frac{\omega}{2 \pi} $

Substitute the values $\left(\omega=88\right.$ and $\left.\pi=\frac{22}{7}\right)$ :

$ \mathrm{f}=\frac{88}{2 \times 22 / 7} $

$\Rightarrow $ $f=\frac{88 \times 7}{44}$

$\Rightarrow $ $\mathrm{f}=14 \mathrm{~Hz}$

When a mass $m$ is attached to a spring with spring constant $k$, the restoring force is given by Hooke's law

$ \mathrm{F}=-\mathrm{kx} $

According to Newton's Second Law,

$ \mathrm{m} \frac{\mathrm{~d}^2 \mathrm{x}}{\mathrm{dt}^2}=-\mathrm{kx} \Rightarrow \frac{\mathrm{~d}^2 \mathrm{x}}{\mathrm{dt}^2}+\frac{\mathrm{k}}{\mathrm{~m}} \mathrm{x}=0 $

Comparing this to the standard SHM equation $\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}+\omega^2 \mathrm{x}=0$, we find the angular frequency as,

$ \omega=\sqrt{\frac{\mathrm{k}}{\mathrm{~m}}} $

Since frequency $v=\frac{\omega}{2 \pi}$, the frequency of oscillation is:

$ v=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{~m}}} $

$\Rightarrow $ $ \mathrm{v} \propto \sqrt{\mathrm{k}} $

A spring constant k is inversely proportional to the natural length L of the spring $\left(\mathrm{k} \propto \frac{1}{\mathrm{~L}}\right)$.

$ \mathrm{k} \cdot \mathrm{~L}=\text { constant } $

Let the initial length be $\mathrm{L}_1$ and initial spring constant be $\mathrm{k}_1$.

The new length is $L_2=\frac{L_1}{2}$.

Let the new spring constant is $\mathrm{k}_2$.

$ \mathrm{k}_1 \mathrm{~L}_1=\mathrm{k}_2 \mathrm{~L}_2 \Rightarrow \mathrm{k}_1 \mathrm{~L}_1=\mathrm{k}_2\left(\frac{\mathrm{~L}_1}{2}\right) $

$\Rightarrow $ $\mathrm{k}_2=2 \mathrm{k}_1$

Using the relation between spring constant and frequency of oscillation,

$ \frac{v_2}{v_1}=\frac{\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_2}{\mathrm{~m}}}}{\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_1}{\mathrm{~m}}}} $

$\Rightarrow $ $ \frac{v_2}{v_1}=\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}} $

Substituting $\mathrm{k}_2=2 \mathrm{k}_1$ :

$ \frac{v_2}{v_1}=\sqrt{\frac{2 \mathrm{k}_1}{\mathrm{k}_1}} $

$\Rightarrow $ $ \frac{v_2}{v_1}=\sqrt{2} $

Therefore, the value of the ratio $\frac{v_2}{v_1}$ is $\sqrt{2}$.

Hence, the correct option is (C).

When the mass is at rest (equilibrium), the downward gravitational force is balanced by the upward spring force.

$ \mathrm{mg}=\mathrm{kx}_0 $

Where, mass of the suspended load is $\mathrm{m}=200 \mathrm{~g}=0.2 \mathrm{~kg}$,

The acceleration due to gravity is $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$,

The extension in spring at equilibrium position is $x_0=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$

Putting the values,

$ \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}_0}=\frac{0.2 \times 10}{2 \times 10^{-3}}=\frac{2}{2} \times 10^{-3}=1000 \mathrm{~N} / \mathrm{m} $

The angular frequency ( $\omega$ ) of a mass-spring system is derived from the restoration force $\mathrm{F}=-\mathrm{kx}$.

Since $\mathrm{F}=\mathrm{ma}$, we have $\mathrm{ma}+\mathrm{kx}=0$,

$ \omega=\sqrt{\frac{\mathrm{k}}{\mathrm{~m}}} $

The linear frequency is $f=\frac{\omega}{2 \pi}$ :

$ \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{1000}{0.2}}=\frac{1}{2 \pi} \sqrt{5000} $

$\Rightarrow $ $ \mathrm{f}=\frac{1}{2 \pi} \sqrt{100 \times 50}=\frac{10 \sqrt{50}}{2 \pi}=\frac{5 \sqrt{50}}{\pi} \mathrm{~Hz} $

The mass is pulled down a further 2 mm from equilibrium and released. This additional displacement is the amplitude (A).

$ A=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m} $

The maximum energy in the spring is the at the point of maximum extension, i.e., $x_{\max }=4 \mathrm{~mm}=4 \times 10^{-3} \mathrm{~m}$

$ \mathrm{E}=\frac{1}{2} \mathrm{kx}_{\max }^2 $

$\Rightarrow $ $E=\frac{1}{2} \times 1000 \times\left(4 \times 10^{-3}\right)^2$

$\Rightarrow $ $ \mathrm{E}=500 \times\left(16 \times 10^{-6}\right)=8000 \times 10^{-6}=8 \times 10^{-3} \mathrm{~J} $

Therefore, the frequency associated with the system is $\frac{5 \sqrt{50}}{\pi} \mathrm{~Hz}$ and the maximum energy in the spring is $8 \times 10^{-3} \mathrm{~J}$.

Thus, the correct option is (A).

For a particle starting from the extreme position $(x=A)$ at $t=0$, the displacement $x$ at any time $t$ is given by:

$ x=A \cos (\omega t) $

Where $\omega$ is the angular frequency, defined as $\omega=\frac{2 \pi}{\mathrm{~T}}$.

The given time period is $\mathrm{T}=5 \mathrm{sec}$ :

$ \omega=\frac{2 \pi}{5} \mathrm{rad} / \mathrm{s} $

The particle moves to $\mathrm{x}=\frac{\mathrm{A}}{\sqrt{2}}$. Substituting into the displacement equation:

$ \frac{\mathrm{A}}{\sqrt{2}}=\mathrm{A} \cos \left(\frac{2 \pi}{5} \mathrm{t}\right) $

$\Rightarrow $ $ \frac{1}{\sqrt{2}}=\cos \left(\frac{2 \pi}{5} t\right) $

Since $\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$ :

$ \frac{\pi}{4}=\frac{2 \pi}{5} \mathrm{t} $

$\Rightarrow $ $ \mathrm{t}=\frac{5}{8} \mathrm{sec} $

Therefore, the correct option is (C).

Let analyse each function one by one.

A. $\sin^2 \omega t$

Using the identity:

$\sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}$

The term $\cos 2\omega t$ has angular frequency $2\omega$, so its time period is:

$T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$

Is it SHM? Although the function oscillates, it is of the form (constant − cosine). The motion has a non-zero mean value ($\frac{1}{2}$), but if we measure displacement from the mean position, $-\frac{1}{2}\cos 2\omega t$ is indeed SHM.

So, A → III (Periodic with $T = \frac{\pi}{\omega}$ and SHM)

B. $\sin^3(2\omega t)$

Using the identity:

$\sin^3 \theta = \frac{3\sin\theta - \sin 3\theta}{4}$

So,

$\sin^3(2\omega t) = \frac{3\sin(2\omega t) - \sin(6\omega t)}{4}$

This is a superposition of two SHMs with periods $\frac{\pi}{\omega}$ and $\frac{\pi}{3\omega}$. The overall period is the LCM, which equals:

$T = \frac{\pi}{\omega}$

Is it SHM? No, because it is a sum of two SHMs of different frequencies (not a single sinusoid).

So, B → I (Periodic with $T = \frac{\pi}{\omega}$ but not SHM)

C. $\sin(\omega t) + \cos(\pi \omega t)$

For the sum to be periodic, the ratio of the two time periods must be a rational number.

• Period of $\sin(\omega t)$: $T_1 = \frac{2\pi}{\omega}$

• Period of $\cos(\pi \omega t)$: $T_2 = \frac{2\pi}{\pi\omega} = \frac{2}{\omega}$

Ratio:

$\frac{T_1}{T_2} = \frac{2\pi/\omega}{2/\omega} = \pi$

Since $\pi$ is irrational, no common period exists. Hence the function is non-periodic.

So, C → IV (Non-periodic)

D. $\cos \omega t + \cos 2\omega t$

• Period of $\cos \omega t$: $T_1 = \frac{2\pi}{\omega}$

• Period of $\cos 2\omega t$: $T_2 = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$

The LCM of $T_1$ and $T_2$ is:

$T = \frac{2\pi}{\omega}$

Is it SHM? No, because it is a superposition of two SHMs of different frequencies.

So, D → II (Periodic with $T = \frac{2\pi}{\omega}$ but not SHM)

Final Matching

$A \to III, \quad B \to I, \quad C \to IV, \quad D \to II$

The correct answer is Option A.

Given,

$ x = a \sin \left(50t + \frac{\pi}{3}\right) $

This is a case of simple harmonic motion, where

$ \omega = 50 \text{ rad/s} $

We need to find:

• $t_1$: time when particle comes to rest, i.e. velocity $v=0$

• $t_2$: time when acceleration is zero, i.e. $a=0$

First, velocity:

$ v = \frac{dx}{dt} = a \cdot 50 \cos\left(50t + \frac{\pi}{3}\right) $

For the particle to come to rest,

$ v=0 $

So,

$ \cos\left(50t + \frac{\pi}{3}\right)=0 $

Now, $\cos \theta = 0$ when

$ \theta = \frac{\pi}{2}, \frac{3\pi}{2}, \dots $

Taking the first time,

$ 50t + \frac{\pi}{3} = \frac{\pi}{2} $

$ 50t = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6} $

$ t_1 = \frac{\pi}{300}\text{ s} $

Now, acceleration in SHM is

$ \alpha = \frac{d^2x}{dt^2} = -\omega^2 x $

So acceleration becomes zero when displacement becomes zero:

$ x=0 $

That means,

$ a\sin\left(50t + \frac{\pi}{3}\right)=0 $

So,

$ \sin\left(50t + \frac{\pi}{3}\right)=0 $

Now, $\sin\theta=0$ when

$ \theta = 0, \pi, 2\pi, \dots $

We take the first positive time. So,

$ 50t + \frac{\pi}{3} = \pi $

$ 50t = \pi - \frac{\pi}{3} = \frac{2\pi}{3} $

$ t_2 = \frac{2\pi}{150} = \frac{\pi}{75}\text{ s} $

Hence,

$ t_1 = \frac{\pi}{300}\text{ s}, \quad t_2 = \frac{\pi}{75}\text{ s} $

So the correct option is:

$ \boxed{\text{Option A}} $

Given:

Natural length of the spring: 2 meters

Initial compression of the spring ($ x_i $): 1 meter

Final compression of the spring when the block is at distance $ x $ ($ x_f $): $ (2 - x) $ meters

Solution:

To find the block's speed, we'll use energy conservation. Initially, the spring's potential energy is fully converted to kinetic energy and spring potential energy at the new position $ x $.

Conservation of Energy:

Initial kinetic energy ($ K_i $) is 0 since the block starts from rest.

Initial potential energy ($ U_i $) due to spring compression is $\frac{1}{2} K x_i^2$.

At a distance $ x $, the kinetic energy ($ K_f $) is $\frac{1}{2} mv^2$ and the potential energy ($ U_f $) is $\frac{1}{2} K x_f^2$.

$ K_i + U_i = K_f + U_f $

$ 0 + \frac{1}{2} K x_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} K x_f^2 $

Rearrange to find $ v^2 $:

$ \frac{1}{2} m v^2 = \frac{1}{2} K (x_i^2 - x_f^2) $

Substitute known values:

$ \frac{1}{2} \times 2 \times v^2 = \frac{1}{2} \times 200 \times \left(1^2 - (2-x)^2 \right) $

Simplify and solve for $ v^2 $:

$ v^2 = 100 \left[ 1 - (2-x)^2 \right] $

Finally, take the square root to find $ v $:

$ v = 10 \left[ 1 - (2-x)^2 \right]^{1/2} $

Therefore, the speed of the block at distance $ x $ from the wall is given by the expression above.

$\begin{aligned} &\text { (A) As both blocks moving together so }\\ &\begin{aligned} & \text { Time period }=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{~K}}} ; \text { where } \mathrm{m}=\mathrm{M}+\mathrm{m} \\ & \qquad \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{~K}}} \end{aligned} \end{aligned}$

(B) Let block is displaced by x in $(+\mathrm{ve})$ direction so force on block will be in(-ve) direction

$\begin{aligned} & F=-K x \\ & (M+m) a=-K x \end{aligned}$

$a=-\frac{K x}{(M+m)}$

(C) As upper block is moving due to friction thus

$\mathrm{f}=\mathrm{ma}=\frac{\mathrm{mKx}}{(\mathrm{M}+\mathrm{m})}$

(D) This option is like two block problem in friction for maximum amplitude, force on block is also maximum, for which both blocks are moving together.

$\begin{aligned} & K A=(M+m) a \\ & a=\frac{K A}{(M+m)} \\ & f=m a=\frac{m K A}{(M+m)} \\ & f_{\max }=f_L=\mu m g \\ & f=\mu m g \\ & \frac{m K A}{(M+m)}=\mu m g \\ & A=\frac{\mu(M+m) g}{K} \end{aligned}$

(E) Maximum friction can be $\mu \mathrm{mg}$ as force is acting between blocks & normal force here is mg .

$\begin{aligned} &\begin{aligned} & x_1=\sqrt{7} \sin 5 t \\ & x_2=2 \sqrt{7} \sin \left(5 t+\frac{\pi}{3}\right) \end{aligned}\\ &\text { From phasor, } \end{aligned}$

$\begin{aligned} &\therefore \text { Amplitude of resultant } \mathrm{SHM}=7\\ &\begin{aligned} & \phi=\tan ^{-1} \frac{2 \sqrt{7} \times \sqrt{3} / 2}{\sqrt{7}+2 \sqrt{7} \times \frac{1}{2}}=\tan ^{-1} \frac{\sqrt{21}}{2 \sqrt{7}}=\tan ^{-1} \frac{\sqrt{3}}{2} \\ & \therefore X_R=7 \sin (5 \mathrm{t}+\phi) \\ & \quad a_R=-7 \times 25 \sin (5 \mathrm{t}+\phi) \\ & \therefore a_{\max }=175 \mathrm{~cm} / \mathrm{sec}=175 \times 10^{-2} \mathrm{~m} / \mathrm{sec} \end{aligned} \end{aligned}$

We know, for an oscillation,

${V_{\max }} = wA$ .... (1)

where, A = amplitude

w = frequency of oscillation

and $w = \sqrt {{k \over m}} $ .... (2)

from (1) and (2), ${V_{\max }} = \sqrt {{k \over m}} A$

$ \Rightarrow {V_{\max }} \propto \sqrt k $ (As m and A are same (or constant))

$ \Rightarrow {{{{\left( {{V_{\max }}} \right)}_A}} \over {{{\left( {{V_{\max }}} \right)}_B}}} = \sqrt {{{{k_1}} \over {{k_2}}}} $

Hence, option 2 is correct.

$\begin{aligned} &\text { As } h \text { increases, } g \text { decreases, } \mathrm{T} \text { increases }\\ &\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}} \\ & \mathrm{~g}=\frac{\mathrm{g}_0 \mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \end{aligned} \end{aligned}$

$\begin{aligned} & \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\phi) \\ & \mathrm{x}_0=\mathrm{A} \sin \phi \quad\text{.... (1)}\\ & \mathrm{p}=\mathrm{mA} \omega \cos (\omega \mathrm{t}+\phi) \\ & \mathrm{p}_0=\mathrm{mA} \omega \cos \phi \quad\text{.... (2)} \end{aligned}$

$(2) /(1) \Rightarrow \tan \phi=\left(\frac{x_0}{p_0}\right) m \omega$

$\sin \phi=\frac{\mathrm{x}_0 \mathrm{~m} \omega}{\sqrt{\left(\mathrm{~m} \omega \mathrm{x}_0\right)^2+\mathrm{p}_0^2}}$

From (1), $A=\frac{x_0}{\sin \phi}=\frac{\sqrt{\left(m \omega x_0\right)^2+p_0^2}}{m \omega}$

This means we can explain assertion with the given reason.

$\begin{aligned} & x=x_0 \sin ^2 \frac{t}{2}=x_0\left(\frac{1-\cos t}{2}\right) \\ & x-\frac{x_0}{2}=\frac{-\cos t}{2} \\ & \text { where } x_0=1 \\ & x-\frac{1}{2}=\frac{-\cos t}{2} \end{aligned}$

Particle is oscillating between $\mathrm{x}=0$ to $\mathrm{x}=1$

$ x(t) = \cos(\pi t) $

The particle executes simple harmonic motion (SHM) with amplitude $ A = 1 \text{ cm} $ and period $ T = 2 \text{ s} $. In one complete cycle (2 s), the motion is as follows:

It starts at $ x = 1 \text{ cm} $.

It moves to $ x = -1 \text{ cm} $ (covering a distance of $ 2 \text{ cm} $).

It returns to $ x = 1 \text{ cm} $ (covering another $ 2 \text{ cm} $).

Thus, the total distance traveled in one complete cycle is:

$ D_{\text{cycle}} = 2 + 2 = 4 \text{ cm}. $

In 12.5 s, the number of cycles is:

$ \frac{12.5}{2} = 6.25 \text{ cycles}. $

For the 6 complete cycles (12 s), the distance traveled is:

$ D_{\text{complete}} = 6 \times 4 = 24 \text{ cm}. $

For the remaining 0.5 s, determine the displacement. At $ t = 12 \text{ s} $, the particle is at:

$ x(12) = \cos(12\pi) = 1 \text{ cm}. $

After an extra 0.5 s (i.e., at $ t = 12.5 \text{ s} $), the position becomes:

$ x(12.5) = \cos(12.5\pi) = \cos\left(12\pi + \frac{\pi}{2}\right) = \cos\frac{\pi}{2} = 0 \text{ cm}. $

The distance covered in this interval is:

$ D_{\text{extra}} = |1 - 0| = 1 \text{ cm}. $

So, the total distance traveled is:

$ D = D_{\text{complete}} + D_{\text{extra}} = 24 + 1 = 25 \text{ cm}. $

The net displacement, which is the difference between the final and initial positions, is calculated as follows:

Initial position at $ t = 0 $:

$ x(0) = \cos 0 = 1 \text{ cm}, $

Final position at $ t = 12.5 \text{ s} $:

$ x(12.5) = 0 \text{ cm}. $

Thus, the displacement is:

$ d = |0 - 1| = 1 \text{ cm}. $

The ratio of the total distance traveled to the displacement is:

$ \frac{D}{d} = \frac{25}{1} = 25. $

Therefore, the correct answer is $25$.

We can determine the time period of the oscillations by considering that when the cube is depressed by a small displacement, the additional buoyant force provided by the displaced water acts as a restoring force. Here’s a step‐by‐step explanation:

Define the given values:

Side length of the cube, $L = 10\text{ cm} = 0.1\text{ m}$.

Mass of the cube, $m = 10\text{ g} = 0.01\text{ kg}$.

Density of water, $\rho = 10^3\ \text{kg/m}^3$.

Acceleration due to gravity, $g = 10\ \text{m/s}^2$.

Cross-sectional area of the cube (face area),

$A = L^2 = (0.1)^2 = 0.01\text{ m}^2.$

When the cube is depressed by a small distance $x$, the additional volume of water displaced is

$\Delta V = A\,x.$

Thus, the additional buoyant force is:

$F_b = \rho\,g\,\Delta V = \rho\,g\,A\,x.$

This force acts in the upward (restoring) direction. Notice that the force is proportional to the displacement $x$, which is the hallmark of simple harmonic motion (SHM).

The effective spring constant, $k$, associated with this SHM is:

$k = \rho\,g\,A.$

The period of oscillation for SHM is given by:

$T = 2\pi \sqrt{\frac{m}{k}},$

which, upon substituting for $k$, becomes:

$T = 2\pi \sqrt{\frac{m}{\rho\,g\,A}}.$

Substitute the given values into the expression:

$T = 2\pi \sqrt{\frac{0.01}{1000 \times 10 \times 0.01}}.$

Calculate the denominator:

$1000 \times 10 \times 0.01 = 100,$

so,

$T = 2\pi \sqrt{\frac{0.01}{100}} = 2\pi \sqrt{0.0001}.$

Since,

$\sqrt{0.0001} = 0.01,$

we have:

$T = 2\pi \times 0.01 = 0.02\pi\ \text{s}.$

The problem states that the time period is given by:

$y \pi \times 10^{-2}\ \text{s}.$

Writing our result in the same form:

$0.02\pi\ \text{s} = 2\pi \times 10^{-2}\ \text{s}.$

We see that $y = 2.$

Therefore, the correct answer is:

Option A: 2.

Assertion (A): A simple pendulum transported to a planet where the mass and radius are 4 times and 2 times that of the Earth, respectively, has the same time period as it does on Earth.

Reason (R): The mass of the pendulum remains unchanged whether on Earth or the other planet.

Explanation:

The acceleration due to gravity $ g $ on a planet is given by the formula:

$ g = \frac{G M}{R^2} $

where $ G $ is the gravitational constant, $ M $ is the mass of the planet, and $ R $ is the radius of the planet.

For the new planet, the gravitational acceleration $ g' $ is:

$ g' = \frac{G(4M)}{(2R)^2} = \frac{4G M}{4R^2} = \frac{G M}{R^2} = g $

Thus, the gravitational acceleration on this new planet is the same as on Earth, $ g $.

The time period $ T $ of a simple pendulum is determined by:

$ T = 2\pi \sqrt{\frac{\ell}{g}} $

where $ \ell $ is the length of the pendulum, which indicates that the time period $ T $ is independent of the mass of the pendulum.

Therefore, while Assertion (A) is true and Reason (R) is true, Reason (R) does not correctly explain Assertion (A) because the time period of the pendulum is also independent of the pendulum's mass, and the key factor is the unchanged gravitational acceleration.

$\begin{aligned} & \frac{\mathrm{ML}^2}{3} \ddot{\theta}+\mathrm{K} \cdot \frac{\ell}{2} \theta \cdot \frac{\ell}{2}+\mathrm{K} \cdot \ell \cdot \theta \cdot \ell=0 \\ & \ddot{\theta}+\frac{15}{4} \frac{\mathrm{k}}{\mathrm{M}} \theta=0 \\ & \ddot{\theta}=-\left(\frac{15}{4} \frac{\mathrm{k}}{\mathrm{M}}\right) \theta \\ & \omega_1=\sqrt{\frac{15 \mathrm{~K}}{4 \mathrm{M}}}\end{aligned}$

$\frac{1}{3} \mathrm{ML}^2 \ddot{\theta}+2 \mathrm{~K} \frac{\mathrm{~L}}{2} \theta \cdot \frac{\mathrm{~L}}{2}=0$

$\begin{aligned} & \ddot{\theta}+\frac{3}{2} \frac{K}{M} \theta=0 \\ & \ddot{\theta}=-\frac{3}{2} \frac{K}{M} \theta \\ & \omega_2=\sqrt{\frac{3}{2} \frac{K}{M}} \\ & \frac{\omega_1}{\omega_2}=\sqrt{\frac{15}{4} \times \frac{2}{3}}=\sqrt{\frac{5}{2}}\end{aligned}$

Since, $F+0.04 \pi^2 y=0$

$ \Rightarrow \quad F=-0.04 \pi^2 y $

For SHM, $F=-K y$

$ \begin{aligned} & \therefore & & K=0.04 \pi^2 \\ & \therefore & \omega & =\sqrt{\frac{K}{m}}=\sqrt{\frac{0.04 \pi^2}{0.09}}=\frac{\sqrt{4 \pi^2}}{9} \\ & \Rightarrow & \omega & =\frac{2 \pi}{3} \mathrm{rad} / \mathrm{s} \\ & \therefore & v_{\max } & =A \omega=\frac{6}{\pi} \times \frac{2 \pi}{3}=4 \mathrm{~m} / \mathrm{s} \end{aligned} $

For a damped oscillator

$ A=A_0 e^{-b t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \text { ( } b=\text { damping constant }) $

So, $A_1=A_0 e^{-b t_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

and $A_2=A_0 e^{-b t_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Multiplying Eqs. (ii) and (iii) we get,

$ \begin{aligned} & A_1 A_2=A_0^2 \cdot e^{-b\left(t_1+t_2\right)} \\ \Rightarrow \quad & \frac{A_1 A_2}{A_0}=A_0 e^{-b\left(t_1+t_2\right)} \end{aligned} $

Comparing with Eq. (i), amplitude at time $\left(t_1+t_2\right)$ is

$ \frac{A_1 A_2}{A_0} $

Step 1: Pendulum Frequency Formula

The time it takes for one swing (period) of a simple pendulum is given by $ T = 2\pi\sqrt{\frac{l}{g}} $, where $ l $ is length and $ g $ is gravity.

Frequency ($ f $) is how many swings it makes in one second, so $ f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{l}} $.

Step 2: Relationship Between Frequency and Length

From this, frequency is inversely related to the square root of length: $ f \propto \frac{1}{\sqrt{l}} $.

This means $ \frac{f_2}{f_1} = \sqrt{\frac{l_1}{l_2}} $, where $ l_1 $ and $ f_1 $ are initial length and frequency, and $ l_2 $ and $ f_2 $ are final.

Step 3: Assign the Values

The number of swings in a minute is frequency per minute. Original frequency = 72, new frequency = 90.

So, $ \frac{f_2}{f_1} = \frac{90}{72} = \frac{5}{4} $.

Now, $ \frac{5}{4} = \sqrt{\frac{l_1}{l_2}} $.

Step 4: Find the New Length

Square both sides: $ (\frac{5}{4})^2 = \frac{l_1}{l_2} $, so $ \frac{25}{16} = \frac{l_1}{l_2} $.

Rearrange to get $ l_2 = \frac{16}{25} l_1 $.

Step 5: Calculate the Percentage Decreased

Decrease in length = $ l_1 - l_2 $.

Percentage decrease = $ \frac{l_1 - l_2}{l_1} \times 100\% $.

Substituting values: $ \frac{l_1 - \frac{16}{25}l_1}{l_1} \times 100\% = \frac{9}{25} \times 100\% = 36\% $.

Understanding the Amplitude Change:

The formula for how the amplitude (A) of a damped oscillator changes over time is: $ A(t) = A_0 e^{-b t / 2 m} $ where A₀ is the starting amplitude, b is the damping constant, m is the mass, and t is time.

We are told the amplitude drops to half its original value in 10 seconds. So, $ A(t) = \frac{A_0}{2} $ when $ t = 10~\text{s} $.

Setting Up the Equation:

We set $ \frac{A_0}{2} = A_0 e^{-10b/2m} $

Divide both sides by A₀ to get: $ \frac{1}{2} = e^{-10b/2m} $

Take the natural logarithm of both sides: $ \ln(2) = \frac{10b}{2m} $

So, $ \frac{b}{m} = \frac{\ln(2)}{5} $  

Now, How Long for Energy to Halve?

Mechanical energy E is proportional to the square of amplitude: $ E \propto A^2 $. It also decays exponentially, but at twice the rate: $ E(t) = E_0 e^{-bt/m} $

We’re looking for time when $ E(t) = \frac{E_0}{2} $

So, $ \frac{E_0}{2} = E_0 e^{-bt/m} $

Divide by E₀: $ \frac{1}{2} = e^{-bt/m} $

Take the natural log: $ \ln(2) = \frac{bt}{m} $

Plug in the Value of $\\frac{b}{m}$:

From before, $ \frac{b}{m} = \frac{\ln(2)}{5} $, so substitute this into the energy equation:

$ \ln(2) = \left[ \frac{\ln(2)}{5} \right] t $

Divide both sides by $ \frac{\ln(2)}{5} $: $ t = 5~\text{seconds} $

Final Answer: It takes 5 seconds for the mechanical energy to become half of its original value.

The force in SHM is $F=-k x$

The maximum force is $F_{\text {max }}=-k A$

Given, $F=0.866 F_{\text {max }}$

$ \begin{aligned}So,\,\, -k x & =0.866(-k A) \\ x & =0.866 \mathrm{~A} \end{aligned} $

The velocity in SHM is $v=\omega \sqrt{\left(A^2-x^2\right)}$

$ \begin{aligned} & v=\omega \sqrt{A^2-(0.866 A)^2} \\ & v=\omega \sqrt{A^2-0.75 A^2}=\omega \sqrt{0.25 A^2} \\ & v=\omega(0.5 A) \end{aligned} $

The maximum velocity in SHM occurs at $x=0$, So $v_{\text {max }}=\omega A$

The ratio is $\frac{v}{v_{\text {max }}}=\frac{0.5 \omega A}{\omega A}$

$ \frac{v}{v_{\max }}=0.5=\frac{1}{2} $

The ratio of its velocity at that point and its maximum velocity is $\frac{1}{2}$.

Given:

$ \frac{U}{K} = \frac{4}{5} $

The amplitude $A = 6\ \mathrm{cm}$

Step 1: Write the energy formulas

The total energy is $E = \frac{1}{2} k A^2$

The potential energy is $U = \frac{1}{2} k x^2$

The kinetic energy is $K = \frac{1}{2} k (A^2 - x^2)$

Step 2: Set up the given ratio

We want the distance $x$ from the mean position so that $\frac{U}{K} = \frac{4}{5}$:

$ \frac{\frac{1}{2}k x^2}{\frac{1}{2}k (A^2 - x^2)} = \frac{4}{5} $

Step 3: Simplify the equation

The $\frac{1}{2}k$ cancels out: $ \frac{x^2}{A^2 - x^2} = \frac{4}{5} $

Step 4: Solve for $x^2$

Multiply both sides by $(A^2 - x^2)$: $ 5 x^2 = 4 (A^2 - x^2) $

Expand: $ 5 x^2 = 4 A^2 - 4 x^2 $

Bring $4x^2$ to the left: $ 5 x^2 + 4 x^2 = 4 A^2 $

Combine like terms: $ 9 x^2 = 4 A^2 $

Step 5: Find $x$

Divide both sides by 9: $ x^2 = \frac{4}{9} A^2 $

Take the square root: $ x = \frac{2}{3} A $

Since $A = 6\ \mathrm{cm}$:

$ x = \frac{2}{3} \times 6 = 4\ \mathrm{cm} $

The distance from the mean position is $4~\mathrm{cm}$.

$ \begin{aligned} & \text { } \sin ^2 \omega t=\frac{1-\cos 2 \omega t}{2}=\frac{1}{2}-\frac{1}{2} \cos 2 \omega t\\ & \therefore \quad T=\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega} \mathrm{s} \end{aligned} $

According to given figure, springs are connected in parallel.$ \begin{aligned} &\text { Therefore effective spring constant }\\ &\begin{aligned} K_{\mathrm{eff}} & =K_1+K_2 \\ & =K+K=2 K \end{aligned} \end{aligned} $

The restoring force on block due to springs not due to gravity pull

So, $\quad T=2 \pi \sqrt{\frac{M}{K_{\text {eff }}}} \Rightarrow T=2 \pi \sqrt{\frac{M}{2 K}}$

$ \begin{aligned} &\text { Displacement, } x=0.5 \cos (125.6 t)\\ &\begin{aligned} \therefore \quad \omega & =125.6 \mathrm{rad} / \mathrm{s} \\ \therefore \quad T & =\frac{2 \pi}{\omega} \\ & =\frac{2 \times 3.14}{125.6} \approx 0.05 \mathrm{~s} \end{aligned} \end{aligned} $

$A=A_0 e^{-b t}$

Since, $A=50 \%$ of $A_0=\frac{A_0}{2}$ at $t=12 \mathrm{~s}$,

$ \begin{array}{ll} \therefore & \frac{A_0}{2}=A_0 e^{-b t} \\ \Rightarrow & 0.5=e^{-12 b} \end{array} $

$ \begin{aligned} &\text { According to second condition, }\\ &\begin{aligned} & A^{\prime}=A_0 e^{-b \times 36}=A_0\left(e^{-12 b}\right)^3=A_0(0.5)^3 \\ & A^{\prime}=0.125 A_0 \\ & A^{\prime}=\frac{A_0}{8} \\ & \Rightarrow \frac{x}{100} A_0=\frac{A_0}{8} \Rightarrow x=12.5 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \frac{k_1}{k_2}= & \frac{\frac{1}{2} m \omega^2\left(A^2-x_1^2\right)}{\frac{1}{2} m \omega^2\left(A^2-x_2^2\right)} \\ & =\frac{A^2-x_1^2}{A^2-x_2^2}=\frac{A^2-\left(\frac{A}{4}\right)^2}{A^2-\left(\frac{A}{2}\right)^2} \\ & =\frac{\frac{15}{16} A^2}{\frac{3}{4} A^2}=\frac{5}{4} \end{aligned} $

For particle in SHM,

$ \begin{aligned} T & =1.5 \mathrm{~s} \mathrm{so} \\ \omega & =\frac{2 \pi}{T}=\frac{4 \pi}{3} \mathrm{rad} / \mathrm{s} \end{aligned} $

Let oscillations starts from $t=0$ and $x=A \sin \omega t$ is displacement equation Then, KE at time $t$ is,

$ \begin{aligned} K & =\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{d x}{d t}\right)^2 \\ & =\frac{1}{2} m \cdot A^2 \omega^2 \cos ^2 \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

And total energy of particle is,

$ E=\frac{1}{2} k A^2 $

Here, $k=m \omega^2$

$ =\frac{1}{2} m A^2 \omega^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} &\text { Given, ratio, } \frac{K}{E}=\frac{3}{4}\\ &\begin{aligned} & \Rightarrow \quad \frac{\frac{1}{2} m A^2 \omega^2 \cos ^2 \omega t}{\frac{1}{2} m A^2 \omega^2}=\frac{3}{4} \\ & \Rightarrow \quad \cos ^2 \omega t=\frac{3}{4} \\ & \Rightarrow \quad \cos \omega t=\sqrt{\frac{3}{2}} \\ & \Rightarrow \quad \omega t=\frac{\pi}{6} \\ & \Rightarrow \quad t=\frac{\pi}{6 \omega}=\frac{\pi \times 3}{6 \times 4 \pi}=\frac{1}{8} \mathrm{~s} \end{aligned} \end{aligned} $

$ \text {} \begin{aligned} y_1 & =0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right) \\ \therefore \quad v_1 & =0.1 \times 100 \pi \cos \left(100 \pi t+\frac{\pi}{3}\right) \\ & =10 \pi \times \cos \left(100 \pi t+\frac{\pi}{3}\right) \\ y_2 & =0.1 \cos \pi t \\ & =0.1 \sin \left(\pi t+\frac{\pi}{2}\right) \\ \therefore \quad v_2 & =0.1 \times \pi \cos \left(\pi t+\frac{\pi}{2}\right) \\ & =0.1 \pi \cos \left(\pi t+\frac{\pi}{2}\right) \end{aligned} $

$ \begin{aligned} \therefore \text { At } t & =0 \\ v_1 & =10 \cos \frac{\pi}{3} \\ \text { and } v_2 & =0.1 \pi \cos \frac{\pi}{2} \end{aligned} $

$\therefore$ Phase difference between $v_1$ and $v_2$ at time $t=0$,

is given as,

$ \phi=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6} $

Step 1: Find Spring Constant ($k$)

The spring stretches by 0.2 m when a 0.5 kg mass hangs from it.

This means the force from the weight ($m_1 g$) equals the force from the spring ($kx$):

$k x = m_1 g$

So,

$k = \frac{m_1 g}{x} = \frac{0.5 \times 10}{0.2} = 25~\mathrm{N}/\mathrm{m}$

Step 2: Use the New Mass

The new hanging mass is $m_2 = 0.25~\mathrm{kg}$.

Step 3: Find the New Time Period ($T$)

The time period of a mass-spring system is:

$T = 2\pi \sqrt{\frac{m_2}{k}}$

Substitute the values:

$T = 2\pi \sqrt{\frac{0.25}{25}}$

$T = 2\pi \sqrt{0.01}$

$T = 2\pi \times 0.1 = 0.2\pi~\mathrm{s}$

Step 4: Calculate the Value

Using $\pi \approx 3.14$:

$T = 0.2 \times 3.14 = 0.628~\mathrm{s}$

• Mass $ m = 4 \, \text{kg} $

• Force constant (spring constant) $ k = 64 \, \text{N/m} $

Formula for time period of a mass–spring system:

$ T = 2\pi \sqrt{\frac{m}{k}} $

Substitute the values:

$ T = 2\pi \sqrt{\frac{4}{64}} = 2\pi \sqrt{\frac{1}{16}} = 2\pi \times \frac{1}{4} = \frac{\pi}{2} $

✅ Answer: $ \boxed{\frac{\pi}{2}\ \text{s}} $

Correct Option: B

Let angular frequency of SHM is $\omega$ and when is at distance $x$ from mean position, its velocity will be given as

$ v=\omega \sqrt{A^2-x^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

When its velocity becomes double, let new amplitude is $A^{\prime}$.

So, $\quad 2 v=\omega \sqrt{A^{\prime 2}-x^2}$

Putting the value of $v$ from equation (i)

$ \begin{aligned} & 2 \omega \sqrt{A^2-x^2}=\omega \sqrt{A^{\prime 2}-x^2} \\ & 2 \sqrt{A^2-x^2}=\sqrt{A^{\prime 2}-x^2} \\ \Rightarrow \quad & 4\left(A^2-x^2\right)=A^{\prime 2}-x^2 \\ \Rightarrow \quad & A^{\prime 2}=4 A^2-3 x^2 \Rightarrow A^{\prime}=\sqrt{4 A^2-3 x^2} \end{aligned} $

$x=A \sin (\omega t+\theta)$

For maximum displacement, $x=A$

$ \begin{array}{ll} & A=A \sin (\omega t+\theta) \\ \Rightarrow & 1=\sin (\omega t+\theta) \\ \Rightarrow & \omega t+\theta=\frac{\pi}{2} \\ \Rightarrow & \omega t=\frac{\pi}{2}-\theta \end{array} $

$ \Rightarrow \quad t=\frac{\pi}{2 \omega}-\frac{\theta}{\omega} $

Given:

• Maximum velocity $ v_{\text{max}} = 5 \, \text{m/s} $

• Maximum acceleration $ a_{\text{max}} = 10 \, \text{m/s}^2 $

For SHM relationships:

$ v_{\text{max}} = \omega A $

$ a_{\text{max}} = \omega^2 A $

where $ \omega $ is the angular frequency, and $ A $ is the amplitude.

Step 1: Divide the two equations

$ \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{\omega^2 A}{\omega A} = \omega $

$ \Rightarrow \omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{10}{5} = 2 \, \text{rad/s} $

Step 2: Find the time period

$ T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \, \text{s} $

✅ Final Answer:

$ \boxed{T = \pi \, \text{s}} $

Correct Option: A) $ \pi \text{ s} $

Combined mass, $m=m_1+m_2$

$ =1+0.5=1.5 \mathrm{~kg} $

Using conservation of momentum,

$ \begin{aligned} & m_1 u_1+m_2 u_2=\left(m_1+m_2\right) v \\ \Rightarrow \quad & 1 \times 0+0.5 \times 3=1.5 v \\ \Rightarrow \quad & v=1 \mathrm{~m} / \mathrm{s} \end{aligned} $

According to conservation of energy,

$ \begin{aligned} & & \frac{1}{2} m v^2 & =\frac{1}{2} k A^2 \\ \Rightarrow & & A & =\sqrt{\frac{m v^2}{k}} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad \sqrt{\frac{1.5 \times 1^2}{600}} & =\sqrt{\frac{1}{400}}=\frac{1}{20} \mathrm{~m} \\ & =\frac{100}{20} \mathrm{~cm}=5 \mathrm{~cm} \end{aligned} $

$ \text { } \begin{aligned} \frac{K}{U} & =\frac{\frac{1}{2} m \omega^2\left(A^2-x^2\right)}{\frac{1}{2} m \omega^2 x^2} \\ & =\frac{A^2-x^2}{x^2} \\ & =\frac{A^2-\left(\frac{A}{2}\right)^2}{\left(\frac{A}{2}\right)^2}\left[\text { Here, } x=\frac{A}{2}\right] \\ & =\frac{\frac{3 A^2}{4}}{\frac{A^2}{4}}=\frac{3}{1} \\ \therefore K: U & =3: 1 \end{aligned} $

$ \begin{aligned} \text { } \quad & T=2 \pi \sqrt{\frac{m}{k}} \\ \frac{T_1}{T_2} & =\sqrt{\frac{m_1}{m_2}} \\ \frac{T}{T+0.2 \pi} & =\sqrt{\frac{4}{9}}=\frac{2}{3} \\ \Rightarrow \quad 3 T & =2 T+0.4 \pi \\ \Rightarrow \quad T & =0.4 \pi \mathrm{~s} \end{aligned} $

$\therefore$ Using Eq. (i),

$ \begin{aligned} & & T & =2 \pi \sqrt{\frac{m}{k}} \\ & \Rightarrow & 0.4 \pi & =2 \pi \sqrt{\frac{4}{k}} \Rightarrow 0.2=\frac{2}{\sqrt{k}} \\ & \Rightarrow & \sqrt{k} & =\frac{1}{0.1} \Rightarrow \sqrt{k}=10 \\ & \Rightarrow & k & =100 \mathrm{Nm}^{-1} \end{aligned} $

Total energy,

$ \begin{aligned} & E=\frac{1}{2} m \omega^2 A^2 \\ & K+U=\frac{1}{2} m \omega^2(0.05)^2 \\ \Rightarrow \quad & 4 \times 10^{-3}+\frac{1}{2} m \omega^2 x^2=\frac{1}{2} m \omega^2(0.05)^2 \\ \Rightarrow \quad & \frac{1}{2} m \omega^2\left[(0.05)^2-x^2\right]=4 \times 10^{-3} \\ \Rightarrow \quad & \frac{1}{2} m \omega^2\left[(0.05)^2-(0.03)^2\right]=4 \times 10^{-3} \\ \Rightarrow \quad & \frac{1}{2} m \omega^2=2.5 \\ \Rightarrow \quad & m \omega^2=5 \end{aligned} $

$\therefore$ Maximum force

$ F_{\max }=m \omega^2 A=5 \times 0.05=0.25 \mathrm{~N} $

$ \begin{aligned} &\text { Angular velocity of the system, }\\ &\begin{aligned} \omega & =\sqrt{\frac{k}{m_{\text {total }}}}=\sqrt{\frac{600}{(1+0.5)}} \\ & =\sqrt{\frac{600}{1.5}}=20 \mathrm{rad} / \mathrm{s} \\ \therefore \quad f & =\frac{\omega}{2 \pi}=\frac{20}{2 \pi}=\frac{10}{\pi} \mathrm{~Hz} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y=5 \sin (3 \pi t)+5 \sqrt{3} \cos (3 \pi t) \\ = & 10\left[\frac{5}{10} \sin 3 \pi t+\frac{5 \sqrt{3}}{10} \cos (3 \pi t)\right] \\ = & 10\left[\sin 3 \pi t \cdot\left(\frac{1}{2}\right)+\cos 3 \pi t \cdot\left(\frac{\sqrt{3}}{2}\right)\right] \\ = & 10\left[\sin (3 \pi t) \cos \frac{\pi}{3}+\cos 3 \pi t \sin \frac{\pi}{3}\right] \\ = & 10 \sin \left(3 \pi t+\frac{\pi}{3}\right) \\ \therefore & A=10 \mathrm{~cm} \end{aligned} \end{aligned} $

$ \begin{aligned} & n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{25}{0.1}} \\ & n=\frac{5}{2 \pi} \mathrm{~Hz} \end{aligned} $

$\therefore$ Angular frequency, $\omega=2 \pi n$

$ =2 \pi \times \frac{5}{2 \pi}=5 \mathrm{rad} \mathrm{~s}^{-1} $

The time period of a simple pendulum is given by the formula:

$T = 2\pi \sqrt{\frac{l}{g}}$

where $T$ is the time period, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity at the location of the pendulum.

The acceleration due to gravity changes with height above the Earth's surface. The acceleration due to gravity at a height $h$ above the Earth's surface can be expressed as:

$g' = g \left(\frac{R}{R + h}\right)^2$

where $g$ is the acceleration due to gravity at the surface of the Earth, $R$ is the radius of the Earth, and $h$ is the height above the Earth’s surface. Since the time period of the pendulum depends on the square root of the inverse of the acceleration due to gravity, any change in $g$ due to a change in height will affect the time period.

Given that the time period of the pendulum at a height $R$ above Earth's surface is $T_1$, and we're to find the time period $T_2$ at a height of $2R$, we can use the formula for acceleration due to gravity at different heights to express the relationship between $T_1$ and $T_2$.

For the initial case at height $R$:

$g_1 = g \left(\frac{R}{R + R}\right)^2 = g \left(\frac{R}{2R}\right)^2 = \frac{g}{4}$

For the new case at height $2R$:

$g_2 = g \left(\frac{R}{R + 2R}\right)^2 = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}$

The time period is proportional to the square root of the inverse of $g$, so:

$\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac{\frac{g}{9}}{\frac{g}{4}}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$

Therefore:

$T_1 = \frac{2}{3}T_2$

Rearranging this equation:

$3T_1 = 2T_2$

This corresponds to Option A.